Question 1 (a)

HOI is a weaker acid than HOCl
Inductive Effects

Question 2 (e)

Percent error is always positive

$Error = Experimental Value- Accepted Value — Can be positive or negative Percent Error = errorl/ accepted value \* 100% — or (experimental value) - (known value) % error = known value — Always positive$

Question 3 (e)

$1 The order of the reaction with respect to C12 is \[Cl\]2 -z. \[Cl\] 1<1/2 \[Cl \]1/2 For step 1, K - \[C121 Substituting into the rate law for step 2 (the slowest step in the mechanism): rate- k\[CH4\] \[Cl\] \[C12\]l/2) \[CH4\] \[C12\]l/2 Because the exponent of C12 in the rate law is 1/2, the order of the reaction with respect to C12 is 1/2. One point is earned for the correct answer with appropriate justification.$

Mechanisms with a Fast Initial Step

$Let's consider an alternative mechanism that does not involve a termolecular step: (fast) NO(g) + Bh(g) Step 1: step 2: NOBh(g) + NO(g) 2NOBr(g) (slow) \[14.27) In this mechanism, step 1 involves two processes: a forward reaction and its reverse. Because step 2 is the rate-determining step, the rate law for that step governs the rate of the overall reaction: Rate = \[14.28) Note that NOBr, is an intermediate generated in the forward reaction of step 1. Inter- mediates are usually unstable and have a low, unknown concentration. Thus, the rate law of Equation 14.28 depends on the unknown concentration of an intermediate, which isn't desirable. We want instead to express the rate law for a reaction in tern-IS of the reactants, or the products if necessary, of the reaction. With the aid of some assumptions, we can express the concentration of the inter- mediate NOBr2 in terms of the concentrations of the starting reactants NO and Br2. We first assume that NOBr, is unstable and does not accumulate to any significant extent in the reaction mixture. Once formed, NOBr2 can be consumed either by reacting with NO to form NOBr or by falling back apart into NO and Br,. The first of these possibili- ties is step 2 of our alternative mechanism, a slow process. The second is the reverse of step 1, a unimolecular process: NO(g) + NOBr2(g) Because step 2 is slow, we assume that most of the NOBr2 falls apart according to this reaction. Thus, we have both the forward and reverse reactions of step 1 occurring much faster than step 2. Because they occur rapidly relative to step 2, the forward and reverse reactions of step 1 establish an equilibrium. As in any other dynamic equilib- rium, the rate of the forward reaction equals that of the reverse reaction: kCNO\]CBr2\] Rate Of forward reaction CNOBr2\] Rate Of reverse r eaction$

$Solving for \[NOBr21, we have \[NOBQ = CNO\]CBr2\] Substituting this relationship into Equation 14.28, we have Rate = = where the experimental rate constant k equals k2k1/k\_1. This expression is consistent with the experimental rate law (Equation 14.25). Thus, our alternative mechanism$